Simplify the following expression and state the condition under which the simplification is valid: $a = \dfrac{q^2 + 2q + 1}{q^2 + 3q + 2}$
First factor the expressions in the numerator and denominator. $ \dfrac{q^2 + 2q + 1}{q^2 + 3q + 2} = \dfrac{(q + 1)(q + 1)}{(q + 2)(q + 1)} $ Notice that the term $(q + 1)$ appears in both the numerator and denominator. Dividing both the numerator and denominator by $(q + 1)$ gives: $a = \dfrac{q + 1}{q + 2}$ Since we divided by $(q + 1)$, $q \neq -1$. $a = \dfrac{q + 1}{q + 2}; \space q \neq -1$